Reduced computation in joint detection

ABSTRACT

A user equipment receives a plurality of transmitted data signals in a communication system. The transmitted data are received. A channel response is determined for each received data signal. A system response is determined based on in part the channel signals. The system response is expanded to be piecewise orthogonal. Data from the received data signals is retrieved based on in part the expanded system response.

CROSS REFERENCE TO RELATED APPLICATIONS

This application is a continuation of U.S. patent application Ser. No. 11/010/703, filed Dec. 13, 2004, which is a continuation of U.S. patent application Ser. No. 09/662,404 filed Sep. 14, 2000, now U.S. Pat. No. 6,831,944, issued Dec. 14, 2004, which in turn claims priority to U.S. Provisional Patent Application No. 60/153,801, filed on Sep. 14, 1999.

BACKGROUND

The invention generally relates to wireless communication systems. In particular, the invention relates to joint detection of multiple user signals in a wireless communication system.

FIG. 1 is an illustration of a wireless communication system 10. The communication system 10 has base stations 12 ₁ to 12 ₅ which communicate with user equipments (UEs) 14 ₁ to 14 ₃. Each base station 12 ₁ has an associated operational area where it communicates with UEs 14 ₁ to 14 ₃ in its operational area.

In some communication systems, such as code division multiple access (CDMA) and time division duplex using code division multiple access (TDD/CDMA), multiple communications are sent over the same frequency spectrum. These communications are typically differentiated by their chip code sequences. To more efficiently use the frequency spectrum, TDD/CDMA communication systems use repeating frames divided into time slots for communication. A communication sent in such a system will have one or multiple associated chip codes and time slots assigned to it based on the communication's bandwidth.

Since multiple communications may be sent in the same frequency spectrum and at the same time, a receiver in such a system must distinguish between the multiple communications. One approach to detecting such signals is single user detection. In single user detection, a receiver detects only the communication from a desired transmitter using a code associated with the desired transmitter, and treats signals of other transmitters as interference.

In some situations, it is desirable to be able to detect multiple communications simultaneously in order to improve performance. Detecting multiple communications simultaneously is referred to as joint detection. Some joint detectors use Cholesky decomposition to perform a minimum mean square error (MMSE) detection and zero-forcing block equalizers (ZF-BLEs). These detectors have a high complexity requiring extensive receiver resources.

Accordingly, it is desirable to have alternate approaches to joint detection.

SUMMARY

A user equipment receives a plurality of transmitted data signals in a communication system. The transmitted data signals are received. A channel response is determined for each received data signal. A system response is determined based on in part the channel signals. The system response is expanded to be piecewise orthogonal. Data from the received data signals is retrieved based on in part the expanded system response.

BRIEF DESCRIPTION OF THE DRAWING(S)

FIG. 1 is a wireless communication system.

FIG. 2 is a simplified transmitter and a receiver using joint detection.

FIG. 3 is an illustration of a communication burst.

FIG. 4 is an illustration of reduced computation joint detection.

DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENT(S)

FIG. 2 illustrates a simplified transmitter 26 and receiver 28 using joint detection in a TDD/CDMA communication system. In a typical system, a transmitter 26 is in each UE 14 ₁ to 14 ₃ and multiple transmitting circuits 26 sending multiple communications are in each base station 12 ₁ to 12 ₅. A base station 12 ₁ will typically require at least one transmitting circuit 26 for each actively communicating UE 14 ₁ to 14 ₃. The joint detection receiver 28 may be at a base station 12 ₁, UEs 14 ₁ to 14 ₃ or both. The joint detection receiver 28 receives communications from multiple transmitters 26 or transmitting circuits 26.

Each transmitter 26 sends data over a wireless communication channel 30. A data generator 32 in the transmitter 26 generates data to be communicated over a reference channel to a receiver 28. Reference data is assigned to one or multiple codes and/or time slots based on the communications bandwidth requirements. A spreading and training sequence insertion device 34 spreads the reference channel data and makes the spread reference data time-multiplexed with a training sequence in the appropriate assigned time slots and codes. The resulting sequence is referred to as a communication burst. The communication burst is modulated by a modulator 36 to radio frequency. An antenna 38 radiates the RF signal through the wireless radio channel 30 to an antenna 40 of the receiver 28. The type of modulation used for the transmitted communication can be any of those known to those skilled in the art, such as direct phase shift keying (DPSK) or quadrature phase shift keying (QPSK).

A typical communication burst 16 has a midamble 20, a guard period 18 and two data bursts 22, 24, as shown in FIG. 3. The midamble 20 separates the two data bursts 22, 24 and the guard period 18 separates the communication bursts to allow for the difference in arrival times of bursts transmitted from different transmitters. The two data bursts 22, 24 contain the communication burst's data and are typically the same symbol length.

The antenna 40 of the receiver 28 receives various radio frequency signals. The received signals are demodulated by a demodulator 42 to produce a baseband signal. The baseband signal is processed, such as by a channel estimation device 44 and a joint detection device 46, in the time slots and with the appropriate codes assigned to the communication bursts of the corresponding transmitters 26. The channel estimation device 44 uses the training sequence component in the baseband signal to provide channel information, such as channel impulse responses. The channel information is used by the joint detection device 46 to estimate the transmitted data of the received communication bursts as soft symbols.

The joint detection device 46 uses the channel information provided by the channel estimation device 44 and the known spreading codes used by the transmitters 26 to estimate the data of the various received communication bursts. Although joint detection is described in conjunction with a TDD/CDMA communication system, the same approach is applicable to other communication systems, such as CDMA.

One approach to joint detection in a particular time slot in a TDD/CDMA communication system is illustrated in FIG. 4. A number of communication bursts are superimposed on each other in the particular time slot, such as K communication bursts. The K bursts may be from K different transmitters. If certain transmitters are using multiple codes in the particular time slot, the K bursts may be from less than K transmitters.

Each data burst 22, 24 of the communication burst 16 has a predefined number of transmitted symbols, such as Ns. Each symbol is transmitted using a predetermined number of chips of the spreading code, which is the spreading factor (SF). In a typical TDD communication system, each base station 12 ₁ to 12 ₅ has an associated scrambling code mixed with its communicated data. The scrambling code distinguishes the base stations from one another. Typically, the scrambling code does not affect the spreading factor. Although the terms spreading code and factor are used hereafter, for systems using scrambling codes, the spreading code for the following is the combined scrambling and spreading codes. Each data burst 22, 24 has Ns×SF chips.

The joint detection device 46 estimates the value that each data burst symbol was originally transmitted. Equation 1 is used to determine the unknown transmitted symbols. r =A d + n   Equation 1

In Equation 1, the known received combined chips, r, is a product of the system response, A, and the unknown transmitted symbols, d. The term, n represents the noise in the wireless radio channel.

For K data bursts, the number of data burst symbols to be recovered is Ns×K. For analysis purposes, the unknown data burst symbols are arranged into a column matrix, d. The d matrix has column blocks, d ₁ to d _(Ns), of unknown data symbols. Each data symbol block, d _(i), has the i^(th) unknown transmitted data symbol in each of the K data bursts. As a result, each column block, d _(i), has K unknown transmitted symbols stacked on top of each other. The blocks are also stacked in a column on top of each other, such that d ₁ is on top of d ₂ and so on.

The joint detection device 46 receives a value for each chip as received. Each received chip is a composite of all K communication bursts. For analysis purposes, the composite chips are arranged into a column matrix, r. The matrix r has a value of each composite chip, totaling Ns*SF chips.

A is the system response matrix. The system response matrix, A, is formed by convolving the impulse responses with each communication burst chip code. The convolved result is rearranged to form the system response matrix, A (step 48).

The joint detection device 46 receives the channel impulse response, h _(i), for each i^(th) one of the K communication bursts from the channel estimation device 44. Each h _(i) has a chip length of W. The joint detection device convolves the channel impulse responses with the known spreading codes of the K communication bursts to determine the symbol responses, s ₁ to s _(K), of the K communication bursts. A common support sub-block, S, which is common to all of the symbol responses is of length K×(SF+W−1).

The A matrix is arranged to have Ns blocks, B₁ to B_(Ns). Each block has all of the symbol responses, s₁ to s_(K) arranged to be multiplied with the corresponding unknown data block in the d matrix, d ₁ to d _(Ns). For example, d ₁ is multiplied with B. The symbol responses, S ₁ to S _(K), form a column in each block matrix, B_(i), with the rest of the block being padded with zeros. In the first block, B₁, the symbol response row starts at the first row. In the second block, the symbol response row is SF rows lower in the block and so on. As a result, each block has a width of K and a height of Ns×SF. Equation 2 illustrates an A block matrix $\begin{matrix} \begin{matrix} {A = \begin{bmatrix} s_{1} & s_{2} & \ldots & s_{K} & 0 & 0 & \ldots & 0 & 0 & 0 & 0 & 0 & \ldots \\ 0 & 0 & 0 & 0 & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \quad \\ \quad & \quad & \quad & \quad & 0 & 0 & 0 & 0 & \quad & \quad & \quad & \quad & \quad \\ \quad & \quad & \quad & \quad & s_{1} & s_{2} & \ldots & s_{K} & \quad & \quad & \quad & \quad & \quad \\ \quad & \quad & \quad & \quad & 0 & 0 & 0 & 0 & \quad & \quad & \quad & \quad & \quad \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & 0 & 0 & 0 & 0 & \ldots \\ \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & s_{1} & s_{2} & \ldots & s_{K} & \quad \\ \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & 0 & 0 & 0 & 0 & \quad \\ \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \vdots & \vdots & \vdots & \vdots & \quad \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \end{bmatrix}} \\ {= \left\lbrack {B_{\quad 1}\quad B_{\quad 2}\quad\ldots\quad B_{\quad N_{\quad s}}} \right\rbrack} \end{matrix} & {{Equation}\quad 2} \end{matrix}$

The n matrix has a noise value corresponding to each received combined chip, totaling Ns×SF chips. For analysis purposes, the n matrix is implicit in the received combined chip matrix, r.

Using the block notation, Equation 1 can be rewritten as Equation 3. $\begin{matrix} \begin{matrix} {r = {{\left\lbrack {B_{\quad 1}\quad B_{\quad 2}\quad B_{3}\quad\ldots\quad B_{\quad N_{\quad s}}} \right\rbrack \times \begin{bmatrix} d_{1} \\ d_{2} \\ d_{3} \\ \vdots \\ d_{N_{s}} \end{bmatrix}} + n}} \\ {= {{\sum\limits_{i = 1}^{N_{s}}{B_{i}d_{i}}} + n}} \end{matrix} & {{Equation}\quad 3} \end{matrix}$

Using a noisy version of the r matrix, the value for each unknown symbol can be determined by solving the equation. However, a brute force approach to solving Equation 1 requires extensive processing. To reduce the processing, the system response matrix, A, is repartitioned. Each block, B_(i) is divided into Ns blocks having a width of K and a height of SF. These new blocks are referred to as A₁ to A_(L) and 0. L is the length of the common support S, as divided by the height of the new blocks, A₁ to A_(L), per Equation 4. $\begin{matrix} {L = \left\lceil \frac{{SF} + W - 1}{SF} \right\rceil} & {{Equation}\quad 4} \end{matrix}$

Blocks A₁ to A_(L) are determined by the supports, s ₁ to s _(K), and the common support, S. A 0 block having all zeros. A repartitioned matrix for a system having a W of 57, SF of 16 and an L of 5 is shown in Equation 5. $\begin{matrix} {A = \begin{bmatrix} A_{1} & 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \\ A_{2} & A_{1} & 0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \\ A_{3} & A_{2} & A_{1} & 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \\ A_{4} & A_{3} & A_{2} & A_{1} & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \\ A_{5} & A_{4} & A_{3} & A_{2} & A_{1} & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \\ 0 & A_{5} & A_{4} & A_{3} & A_{2} & A_{1} & 0 & \cdots & 0 & 0 & 0 & 0 \\ 0 & 0 & A_{5} & A_{4} & A_{3} & A_{2} & A_{1} & ⋰ & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & A_{5} & A_{4} & A_{3} & A_{2} & ⋰ & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & A_{5} & A_{4} & A_{3} & ⋰ & A_{1} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & A_{5} & A_{4} & ⋰ & A_{2} & A_{1} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & A_{5} & ⋰ & A_{3} & A_{2} & A_{1} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & ⋰ & A_{4} & A_{3} & A_{2} & A_{1} \end{bmatrix}} & {{Equation}\quad 5} \end{matrix}$

To reduce the complexity of the matrix, a piecewise orthogonalization approach is used. Any of the blocks B_(i) for i being L or greater is non-orthogonal to any of the preceding L blocks and orthogonal to any blocks preceding by more than L. Each 0 in the repartitioned A matrix is an all zero block. As a result to use a piecewise orthogonalization, the A matrix is expanded (step 50).

The A matrix is expanded by padding L−1 zero blocks to the right of each block of the A matrix and shifting each row in the A matrix by its row number less one. To illustrate for the A1 block in row 2 of FIG. 2, four (L−1) zeros are inserted between A2 and A1 in row 2. Additionally, block A1 (as well as A2) is shifted to the right by one column (row 2-1). As a result, Equation 5 after expansion would become Equation 6. $\begin{matrix} {A_{\exp} = \left\lbrack \quad\begin{matrix} A_{1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad \\ 0 & A_{2} & 0 & 0 & 0 & A_{1} & 0 & 0 & 0 & 0 & 0 & \quad \\ 0 & 0 & A_{3} & 0 & 0 & 0 & A_{2} & 0 & 0 & 0 & A_{1} & \quad \\ 0 & 0 & 0 & A_{4} & 0 & 0 & 0 & A_{3} & 0 & 0 & 0 & \quad \\ 0 & 0 & 0 & 0 & A_{5} & 0 & 0 & 0 & A_{4} & 0 & 0 & \quad \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & A_{5} & 0 & \ldots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad \end{matrix}\quad \right\rbrack} & {{Equation}\quad 6} \end{matrix}$

To accommodate the expanded A matrix, the d matrix must also be expanded, d _(exp). Each block, d ₁ to d _(Ns), is expanded to a new block, d _(exp1) to d _(expNs). Each expanded block, d _(exp1) to d _(expNs), is formed by repeating the original block L times. For example for d _(exp1) a first block row would be created having L versions of d1, stacked one below the other.

As a result, Equation 1 can be rewritten as Equation 7. $\begin{matrix} \begin{matrix} {r = {{A_{\exp} \cdot d_{\exp}} + n}} \\ {= {{\left\lbrack {B_{\exp\quad 1}\quad B_{\exp\quad 2}\quad B_{\exp\quad 3}\quad\ldots\quad B_{\exp\quad N_{s}}} \right\rbrack \times \begin{bmatrix} d_{\exp\quad 1} \\ d_{\exp\quad 2} \\ d_{\exp\quad 3} \\ \vdots \\ d_{\exp\quad N_{s}} \end{bmatrix}} + n}} \\ {{= {{\sum\limits_{i = 1}^{N_{s}}{B_{\exp\quad i}d_{\exp\quad i}}} + n}},} \end{matrix} & {{Equation}\quad 7} \end{matrix}$ Equation 7 can be rewritten to partition each B_(expi) orthogonally in L partitions, U_(j) ^((I)), j=1 to L, as in Equation 8. $\begin{matrix} \begin{matrix} {r = {{A_{\exp} \cdot d_{\exp}} + n}} \\ {= {{\sum\limits_{i = 1}^{N_{s}}{\left\lbrack {U_{1}^{(i)}\quad U_{2}^{(i)}\quad\ldots\quad U_{L}^{(i)}} \right\rbrack \times \begin{bmatrix} d_{i} \\ d_{i} \\ d_{i} \\ \vdots \\ d_{i} \end{bmatrix}}} + n}} \\ {= {{\sum\limits_{i = 1}^{N_{s}}{\sum\limits_{j = 1}^{L}{U_{j}^{(i)}d_{i}}}} = {{\sum\limits_{i = 1}^{N_{s}}{B_{i}d_{i}}} + n}}} \end{matrix} & {{Equation}\quad 8} \end{matrix}$

To reduce computational complexity, a QR decomposition of the A_(exp) matrix is performed (step 52). Equation 9 illustrates the QR decomposition of A_(exp). A_(exp)=Q_(exp)R_(exp)  Equation 9 Due to the orthogonal partitioning of A_(exp), the QR decomposition of A_(exp) is less complex. The resulting Q_(exp) and R_(exp) matrices are periodic with an initial transient extending over L blocks. Accordingly, Q_(exp) and R_(exp) can be determined by calculating the initial transient and one period of the periodic portion. Furthermore, the periodic portion of the matrices is effectively determined by orthogonalizing A₁ to A_(L). One approach to QR decomposition is a Gramm-Schmidt orthogonalization.

To orthogonalize A_(exp) as in Equation 6, B_(exp1) is orthogonalized by independently orthogonalizing each of its orthogonal partitions, {U_(j) ^((i))}, j=1 . . . L. Each {A_(j)}, j=1 . . . L is independently orthogonalized, and the set is zero-padded appropriately. {Q_(j)} are the orthonormal sets obtained by orthogonalizing {U_(j) ^((i))}. To determine B_(exp2), its U₁ ⁽²⁾ needs to be orthogonalized with respect to only Q₂ of B_(exp1) formed previously. U₂ ⁽²⁾, U₂ ⁽²⁾ and U₄ ⁽²⁾ only need to be orthogonalized with respect to only Q₃, Q₄ and Q₅, respectively. U₅ ⁽²⁾ needs to be orthogonalized to all previous Qs and its orthogonalized result is simply a shifted version of Q₅ obtained from orthogonalizing B_(exp1).

As the orthogonalizing continues, beyond the initial transient, there emerges a periodicity which can be summarized as follows. The result of orthogonalizing B_(expi), i≧6 can be obtained simply by a periodic extension of the result of orthogonalizing B_(exp5).

The orthogonalization of B_(exp5), is accomplished as follows. Its Q₅ is obtained by orthogonalizing A₅, and then zero padding. Its Q₄ is obtained by orthogonalizing the support of Q₅ and A₄, [sup(Q₅) A₄], and then zero padding. Since sup(Q₅) is already an orthogonal set, only A₄ needs to be orthogonalized with respect to sup(Q₅) and itself. Its Q₃ is obtained by orthogonalizing [sup(Q₅) sup(Q₄) A₃] and then zero padding. Its Q₂ is obtained by orthogonalizing [sup(Q₅) sup(Q₄) sup(Q₃) A₂] and then zero padding. Its Q. is obtained by orthogonalizing [sup(Q₅) sup(Q₄) sup(Q₃) sup(Q₂) A₁] and then zero padding. Apart from the initial transient, the entire A_(exp) can be efficiently orthogonalized, by just orthogonalizing A_(p) per Equation 10. A_(p)=[A₅A₄A₃A₂A₁]  Equation 10 By effectively orthogonalizing the periodic portion of A_(exp) by using only A_(p), computational efficiency is achieved. Using a more compact notation, Q_(i) ^(s),for sup (Q_(i)), this orthogonalization of A_(p) results in the orthonormal matrix, Q_(p), of Equation 11. Q_(p)=[Q₅ ^(S)Q₄ ^(S)Q₃ ^(S)Q₂ ^(S)Q₁ ^(S)]  Equation 11 The periodic part of Q_(exp) is per Equation 12. $\begin{matrix} {{PeriodicPartofQ}_{\exp} = \left\lbrack \quad\begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \quad \\ Q_{1}^{S} & 0 & 0 & 0 & 0 & \quad & 0 & 0 & 0 & 0 & \quad \\ 0 & Q_{2}^{S} & 0 & 0 & 0 & Q_{1}^{S} & 0 & 0 & 0 & 0 & \quad \\ 0 & 0 & Q_{3}^{S} & 0 & 0 & 0 & Q_{2}^{S} & 0 & 0 & 0 & \quad \\ 0 & 0 & 0 & Q_{4}^{S} & 0 & 0 & 0 & Q_{3}^{S} & 0 & 0 & \quad \\ 0 & 0 & 0 & 0 & Q_{5}^{S} & 0 & 0 & 0 & Q_{4}^{S} & 0 & \ldots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & Q_{5}^{S} & \quad \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & 0 & \quad \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \end{matrix}\quad \right\rbrack} & {{Equation}\quad 12} \end{matrix}$

To constructing the upper triangular matrix R_(exp),

A_(i)

_(j) is a block of size K×K representing the projections of each column of A_(i) onto all the columns of Q_(j) ^(s). For example, the first column of

A₄

₅ represents the projections of the first column of A₄ on each of the K columns of Q₅ ^(s). Similarly,

A₄

₄ represents the projections of the first column of A₄ on each of the K columns of Q₄ ^(S). However, this block will be upper triangular, because the k^(th) column of A₄ belongs to the space spanned by the orthonormal vectors of Q₅ ^(S) and the first k vectors of Q₄ ^(S). This block is also orthogonal to subsequent vectors in Q₄ ^(S), leading to an upper triangular

A₄

₄. Any

A_(i)

_(j) with i=j will be upper triangular. To orthogonalize other blocks, the following results.

The first block of B_(exp5), viz., U₁ ⁽⁵⁾ is formed by a linear combination of {Q_(j) ^(S)}, j=1 . . . 5, with coefficients given by

A₁

_(j), j=1 . . . 5. The second block, U₂ ⁽⁵⁾, is formed by a linear combination of {Q_(j) ^(S)}, j=2 . . . 5, with coefficients given by

A₂

_(j), j=2 . . . 5. The third block, U₃ ⁽⁵⁾, is formed by a linear combination of {Q_(j) ^(S)}, j=3 . . . 5, with coefficients given by

A₂

_(j), j=3 . . . 5. The fourth block, U₄ ⁽⁵⁾,is formed by a linear combination of {Q_(j) ^(S)}, j=4, 5, with coefficients given by

A₂

_(j), j=4, 5. The fifth block, U₅ ⁽⁵⁾, is formed by Q₅ ^(S)×

A₅

₅.

Accordingly, the coefficients in the expansion of subsequent B_(expi), i≧6 are simply periodic extensions of the above. Since the R_(exp) entries are computed during the orthogonalization of A_(exp), no additional computations are needed to construct R_(exp). Disregarding the initial transient, the remainder of R_(exp) is periodic, and two periods of it are shown in Equation 13. $\begin{matrix} {R_{\exp} = \begin{bmatrix} \quad & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\quad\ldots} \\ \quad & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \quad & \quad \\ \quad & {0\quad} & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad \\ \quad & {\left\langle \quad A_{\quad 1} \right\rangle_{5}\quad} & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad \\ \quad & {0\quad} & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad \\ \quad & {0\quad} & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad \\ \quad & 0 & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad & \quad \\ \quad & {\left\langle \quad A_{\quad 1} \right\rangle_{4}\quad} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad & \quad \\ \quad & 0 & {\left\langle \quad A_{\quad 2} \right\rangle_{5}\quad} & 0 & 0 & 0 & {\left\langle \quad A_{\quad 1} \right\rangle_{5}\quad} & 0 & 0 & 0 & \quad & \quad \\ \quad & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad & \quad \\ \quad & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad & \quad \\ \quad & {\left\langle \quad A_{\quad 1} \right\rangle_{3}\quad} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad & \quad \\ \quad & 0 & {\left\langle \quad A_{\quad 2} \right\rangle_{4}\quad} & 0 & 0 & 0 & {\left\langle \quad A_{\quad 1} \right\rangle_{4}\quad} & 0 & 0 & 0 & \quad & \quad \\ \quad & 0 & 0 & {\left\langle \quad A_{\quad 3} \right\rangle_{5}\quad} & 0 & 0 & 0 & {\left\langle \quad A_{\quad 2} \right\rangle_{5}\quad} & 0 & 0 & \quad & \quad \\ \quad & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad & \quad \\ \quad & {\left\langle \quad A_{\quad 1} \right\rangle_{2}\quad} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad & \quad \\ 0 & 0 & {\left\langle \quad A_{\quad 2} \right\rangle_{3}\quad} & 0 & 0 & 0 & {\left\langle \quad A_{\quad 1} \right\rangle_{3}\quad} & 0 & 0 & 0 & \quad & \quad \\ \quad & 0 & 0 & {\left\langle \quad A_{\quad 3} \right\rangle_{4}\quad} & 0 & 0 & 0 & {\left\langle \quad A_{\quad 2} \right\rangle_{4}\quad} & 0 & 0 & \quad & \quad \\ \quad & 0 & 0 & 0 & {\left\langle \quad A_{\quad 4} \right\rangle_{5}\quad} & 0 & 0 & 0 & {\left\langle \quad A_{\quad 3} \right\rangle_{5}\quad} & 0 & \quad & {\ldots\quad} \\ \quad & {\left\langle \quad A_{\quad 1} \right\rangle_{1}\quad} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \quad & \quad \\ \quad & 0 & {\left\langle \quad A_{\quad 2} \right\rangle_{2}\quad} & 0 & 0 & 0 & {\left\langle \quad A_{\quad 1} \right\rangle_{2}\quad} & 0 & 0 & 0 & \quad & \quad \\ \quad & 0 & 0 & {\left\langle \quad A_{\quad 3} \right\rangle_{3}\quad} & 0 & 0 & 0 & {\left\langle \quad A_{\quad 2} \right\rangle_{3}\quad} & 0 & 0 & \quad & \quad \\ \quad & 0 & 0 & 0 & {\left\langle \quad A_{\quad 4} \right\rangle_{4}\quad} & 0 & 0 & 0 & {\left\langle \quad A_{\quad 3} \right\rangle_{4}\quad} & 0 & \quad & \quad \\ \quad & 0 & 0 & 0 & 0 & {\left\langle \quad A_{\quad 5} \right\rangle_{5}\quad} & 0 & 0 & 0 & {\left\langle \quad A_{\quad 4} \right\rangle_{5}\quad} & \quad & \quad \\ \quad & 0 & 0 & 0 & 0 & 0 & {\left\langle \quad A_{\quad 1} \right\rangle_{1}\quad} & 0 & 0 & 0 & \quad & \quad \\ \quad & 0 & 0 & 0 & 0 & 0 & 0 & {\left\langle \quad A_{\quad 2} \right\rangle_{2}\quad} & 0 & 0 & \quad & \quad \\ \quad & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\left\langle \quad A_{\quad 3} \right\rangle_{3}\quad} & 0 & \quad & \quad \\ \quad & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\left\langle \quad A_{\quad 4} \right\rangle_{4}\quad} & \quad & \quad \\ \quad & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\left\langle \quad A_{\quad 5} \right\rangle_{5}\quad} & \quad \\ \quad & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \left\langle \quad A_{\quad 1} \right\rangle_{1} \end{bmatrix}} & {{Equation}\quad 16} \end{matrix}$ The least squares approach to solving Q_(exp) and R_(exp) is shown in Equation 14. Q _(exp) ·R _(exp) ·d _(exp)= r   Equation 14 By pre-multiplying both sides of Equation 14 by the transpose of Q_(exp), Q_(exp) ^(T), and using Q_(exp) ^(T)·Q_(exp)=I_(LKN) _(s) , Equation 14 becomes Equation 15. R _(exp) ·d_(exp) = Q _(exp) ^(T) r   Equation 15 Equation 15 represents a triangular system whose solution also solves the LS problem of Equation 14.

Due to the expansion, the number of unknowns is increased by a factor of L. Since the unknowns are repeated by a factor of L, to reduce the complexity, the repeated unknowns can be collected to collapse the system. R_(exp) is collapsed using L coefficient blocks, CF₁ to CF_(L), each having a width and a height of K. For a system having an L of 5, CF₁ to CF₅ can be determined as in Equation 16. CF ₁=

A ₁

₁+

A ₂

₂+

A ₃

₃+

A ₄

₄+

A ₅

₅ CF ₂=

A ₁

₂+

A ₂

₃+

A ₃

₄+

A ₄

₅ CF ₃=

A ₁

₃+

A ₂

₄+

A ₃

₅ CF ₄=

A ₁

₄+

A ₂

₅ CF₅=

A₁

₅  Equation 16 Collapsing R_(exp) using the coefficient blocks produces a Cholesky like factor, Ĝ (step 54). By performing analogous operations on the right hand side of Equation 15 results in a banded upper triangular system of height and width of K×Ns as in Equation 17. $\begin{matrix} {{\left\lbrack \quad\begin{matrix} {Tr}_{1} & {Tr}_{2} & {Tr}_{3} & {Tr}_{4} & {CF}_{5} & 0 & 0 & 0 & 0 & 0 & \ldots \\ 0 & {Tr}_{1} & {Tr}_{2} & {CF}_{3} & {CF}_{4} & {CF}_{5} & 0 & 0 & 0 & 0 & \ldots \\ 0 & 0 & {CF}_{1} & {CF}_{2} & {CF}_{3} & {CF}_{4} & {CF}_{5} & 0 & 0 & 0 & \ldots \\ 0 & 0 & 0 & {CF}_{1} & {CF}_{2} & {CF}_{3} & {CF}_{4} & {CF}_{5} & 0 & 0 & \ldots \\ 0 & 0 & 0 & 0 & {CF}_{1} & {CF}_{2} & {CF}_{3} & {CF}_{4} & {CF}_{5} & 0 & \ldots \\ \vdots & \vdots & \vdots & \vdots & 0 & ⋰ & ⋰ & ⋰ & ⋰ & ⋰ & 0 \end{matrix} \right\rbrack \times \left\lbrack \quad\begin{matrix} d_{1} \\ d_{2} \\ d_{3} \\ \vdots \\ d_{N_{s}} \end{matrix} \right\rbrack} = \quad\hat{r}} & {{Equation}\quad 17} \end{matrix}$ Tr₁ to Tr₄ are the transient terms and {circumflex over (r)}. By solving the upper triangle via back substitution, Equation 17 can be solved to determine d (step 56). As a result, the transmitted data symbols of the K data bursts is determined.

Using the piecewise orthogonalization and QR decomposition, the complexity of solving the least square problem when compared with a banded Cholesky decomposition is reduced by a factor of 6.5. 

1. A circuit for receiving a plurality of transmitted data signals, the circuit comprising: an antenna for receiving the transmitted data signals; a channel estimation device for determining a channel response for each received data signal; and a joint detection device having an input configured to receive the channel responses and the received data signals for determining a system response based on in part the channel signals, expanding the system response to be piecewise orthogonal, and retrieving data from the received data signals based on in part the expanded system response.
 2. The circuit of claim 1 for use in a time division duplex using code division multiple access communication system.
 3. The circuit of claim 2 wherein each of the transmitted data signals has an associate code and is transmitted in a shared frequency spectrum and the system response is determined by convolving associated chip codes with the channel response.
 4. The circuit of claim 2 wherein the channel estimation device measures the channel response using a received training sequence associated with the data signals.
 5. The circuit of claim 1 wherein the system response is a system response matrix, further comprising dividing the system response matrix into blocks of columns prior to the expanding.
 6. The circuit of claim 5 wherein the expanding is by padding zeros in the column blocks such that each column block is orthogonal. 